Q3..Consider a system in which a process is allocated three frames in memory.Given the page requests of this process as "1, 4, 5, 3, 4, 5, 6, 1, 7, 5, 6, 1, 3, 2, 6, 5, 2, 1, 5, 2",trace the utilization of the three frames by this process when using the Optimal, FIFO, LRUreplacement algorithms. To compare the efficiency of using each algorithm, find the totalnumber of page faults and the effective access time. Assume that the memory access timeis 100 nanoseconds and the average page-fault service time is 4 microseconds.012012012Optimal Replacement1453 456 175 61326 52 152Number of page faults =Effective Access Time =14 5 3 4 5 6Number of page faults=Effective Access Time =Number of page faults =FIFO ReplacementEffective Access Time =175613265215LRU Replacement1453 456 1756 1 3 2 6 521 522

optimal replacement:- number of hit=9hit ratio =920=45%number of page fault=11fault…

Answered: Q3. Consider a system in which a…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

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Q3..
Consider a system in which a process is allocated three frames in memory.
Given the page requests of this process as "1, 4, 5, 3, 4, 5, 6, 1, 7, 5, 6, 1, 3, 2, 6, 5, 2, 1, 5, 2",
trace the utilization of the three frames by this process when using the Optimal, FIFO, LRU
replacement algorithms. To compare the efficiency of using each algorithm, find the total
number of page faults and the effective access time. Assume that the memory access time
is 100 nanoseconds and the average page-fault service time is 4 microseconds.
0
1
2
0
12
0
1
2
Optimal Replacement
1453 456 175 61326 52 152
Number of page faults =
Effective Access Time =
14 5 3 4 5 6
Number of page faults=
Effective Access Time =
Number of page faults =
FIFO Replacement
Effective Access Time =
175613265215
LRU Replacement
1453 456 1756 1 3 2 6 521 52
2

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Step 1

optimal replacement:-

$number of hit = 9 hit ratio = \frac{9}{20} = 45 % number of page fault = 11 fault ratio = \frac{11}{20} = 55 % = 0.55 effective access time 0 \cdot 55 \times 4000 + (1 - 0 \cdot 55) \times 100 nanosecond 2200 + 45 2245 ns$

FIFO replacement:-

$number of hit = 4 hit ratio = \frac{4}{20} = 20 % number of page fault = 16 fault ratio = \frac{16}{20} = 80 % = 0 \cdot 8 effective access time 0 \cdot 8 \times 4000 + (1 - 0 \cdot 8) \times 100 nanosecond 3200 + 20 3220 n s$

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